`int (dx)/((x+1)^(2)(x^(2)+1))`

Explanation

To solve the integral \( \int \frac{dx}{(x+1)^2 (x^2+1)} \), we will use the method of partial fractions. Here’s the step-by-step solution:

### Step 1: Set up the partial fraction decomposition
We start by expressing the integrand as a sum of simpler fractions. We can write:
\[ \frac{1}{(x+1)^2 (x^2+1)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{Cx + D}{x^2 + 1} \]
where \(A\), \(B\), \(C\), and \(D\) are constants to be determined.

### Step 2: Clear the denominators
Multiply both sides by \((x+1)^2 (x^2+1)\) to eliminate the denominators:
\[ 1 = A(x+1)(x^2+1) + B(x^2+1) + (Cx + D)(x+1)^2 \]

### Step 3: Expand the right-hand side
Now we expand the right-hand side:
\[ 1 = A(x^3 + x^2 + x + 1) + B(x^2 + 1) + (Cx + D)(x^2 + 2x + 1) \]
Expanding \((Cx + D)(x^2 + 2x + 1)\) gives:
\[ Cx^3 + 2Cx^2 + Cx + Dx^2 + 2Dx + D \]
Combining all terms, we have:
\[ 1 = (A + C)x^3 + (A + B + 2C + D)x^2 + (A + C + 2D)x + (A + B + D) \]

### Step 4: Set up equations for coefficients
Since the left-hand side is \(1\) (which has no \(x\) terms), we can equate coefficients:
1. \(A + C = 0\) (coefficient of \(x^3\))
2. \(A + B + 2C + D = 0\) (coefficient of \(x^2\))
3. \(A + C + 2D = 0\) (coefficient of \(x\))
4. \(A + B + D = 1\) (constant term)

### Step 5: Solve the system of equations
From equation (1), we have \(C = -A\). Substituting \(C = -A\) into the other equations:
- From (2): \(A + B + 2(-A) + D = 0 \Rightarrow B - A + D = 0 \Rightarrow D = A - B\)
- From (3): \(A + (-A) + 2D = 0 \Rightarrow 2D = 0 \Rightarrow D = 0\)
- From (4): \(A + B + 0 = 1 \Rightarrow A + B = 1\)

Now substituting \(D = 0\) into \(D = A - B\):
\[ 0 = A - B \Rightarrow A = B \]
Substituting \(A = B\) into \(A + B = 1\):
\[ 2A = 1 \Rightarrow A = \frac{1}{2}, B = \frac{1}{2} \]
And from \(C = -A\):
\[ C = -\frac{1}{2} \]

### Step 6: Write the partial fraction decomposition
Thus, we have:
\[ \frac{1}{(x+1)^2 (x^2+1)} = \frac{1/2}{x+1} + \frac{1/2}{(x+1)^2} - \frac{1/2}{x^2 + 1} \]

### Step 7: Integrate each term
Now we can integrate term by term:
\[ \int \frac{dx}{(x+1)^2 (x^2+1)} = \frac{1}{2} \int \frac{dx}{x+1} + \frac{1}{2} \int \frac{dx}{(x+1)^2} - \frac{1}{2} \int \frac{dx}{x^2 + 1} \]

Calculating each integral:
1. \(\int \frac{dx}{x+1} = \ln|x+1|\)
2. \(\int \frac{dx}{(x+1)^2} = -\frac{1}{x+1}\)
3. \(\int \frac{dx}{x^2 + 1} = \tan^{-1}(x)\)

### Step 8: Combine the results
Putting it all together:
\[ \int \frac{dx}{(x+1)^2 (x^2+1)} = \frac{1}{2} \ln|x+1| - \frac{1}{2(x+1)} - \frac{1}{2} \tan^{-1}(x) + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{dx}{(x+1)^2 (x^2+1)} = \frac{1}{2} \ln|x+1| - \frac{1}{2(x+1)} - \frac{1}{2} \tan^{-1}(x) + C \]